The two possible positions of an object kept in front of a lens of power +5.0 D, so that the image formed in both cases is four times magnified, are

(A) -30 cm
(B) -25 cm
(C) -15 cm
(D) -45 cm

Choose the correct answer from the options given below:

(B) and (C) only

The correct answer is Option (4) → (B) and (C) only

Given:

Lens power: $P = +5.0\ \text{D} \Rightarrow f = \frac{100}{5} = 20\ \text{cm}$

Magnification: $|m| = 4$

Lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$, and magnification $m = \frac{v}{u}$

For $m = +4$ (virtual, erect image): $v = 4u$

$\frac{1}{20} = \frac{1}{4u} - \frac{1}{u} = \frac{1 - 4}{4u} = \frac{-3}{4u} \Rightarrow u = -15\ \text{cm}$

For $m = -4$ (real, inverted image): $v = -4u$

$\frac{1}{20} = \frac{-1}{4u} - \frac{1}{u} = \frac{-1 - 4}{4u} = \frac{-5}{4u} \Rightarrow u = -25\ \text{cm}$

∴ Two object positions: -15 cm and -25 cm